Suppose that the minimum possible value is $a$, and the maximum possible value is $b$. We want to map the interval $[a,b]$ to the interval $[0,200]$.

There are a great many possibilities. One of the simplest is a linear mapping, where $x$ is mapped to $f(x)$, with $f(x)=px+q$.

We want $f(a)=0$, so $pa+q=0$. We want $f(b)=200$, so $pb+q=200$.

Now solve the system of equations $pa+q=0$, $\,pb+q=200$ for $p$ and $q$. We get
$$p=\frac{200}{b-a},\qquad q=-\frac{200a}{b-a},$$
and therefore
$$f(x)=\frac{200}{b-a}\left(x-a\right).$$
You may want to modify this so that the scaled values are integers. If so, then a sensible modification is to use, instead of $f(x)$, the function $g(x)$, where $g(x)$ is the nearest integer to $f(x)$.

Added: If you want to have only $N$ values, equally spaced with bottom at $0$ and top at $200$, then $N-1$ should divide $200$. Let $D=\frac{200}{N-1}$. Take the integer nearest to $\frac{f(x)}{D}$, and multiply the result by $D$. For example, if you want $26$ different values $0, 8, 16, 24, \dots, 200$, then $N=26$, so $N-1=25$ and $D=8$. We find the nearest integer to $\frac{f(x)}{8}$ and multiply the result by $8$.

Remark: One important real world example is when you want to convert the temperature $x$, in degrees Fahrenheit, to temperature in degrees Celsius. In the Fahrenheit scale, the freezing point of water is $32$ degrees, and the boiling point is $212$. In degrees Celsius it is $0$ and $100$. So we want to transform the interval $[32,212]$ to the interval $[0,100]$. the relevant $f(x)$ is equal to $\frac{100}{212-32}(x-32)$, which simplifies to $f(x)=\frac{5}{9}(x-32)$.

Thanks @Andre for your nice explanation. please review the updated question modification and suggest me about that.. i want equal intervals between min and max values. i know only min and max value, not mapping some external scale. i am not doing exact that done in link specified in my question.
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Niranjan KalaJun 19 '12 at 5:03

Min and max look like all you need. If you only want $N$ equally spaced values, like $N=11$ (we want $N-1$ to divide $200$, the final rounding should be a little different from the one I described. I will add something about that to the post.
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André NicolasJun 19 '12 at 5:16

+1:thanks @Andre for reply. i will check and wait for your edits..
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Niranjan KalaJun 19 '12 at 5:23

@NiranjanKala: The first time I wrote out the modification, I made a mistake. It is now right.
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André NicolasJun 19 '12 at 5:29

well, it's alright i will try to check @Gerry answer and your will also help others.. have a nice time..
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Niranjan KalaJun 19 '12 at 5:35